Step 1–Generalized eigenvalue decomposition. 
We start by applying generalized eigenvalue decomposition
[1]} to the positive define matrices \(\mathbf {B}^*+\mathbf {N}_1\)  and
\(\mathbf {B}^*+\mathbf {N}_2\) . There exists an invertible generalized
eigenvector matrix \(\mathbf {V}\)  such that
\(& \mathbf {V}^{(\mathbf {B}^*+\mathbf {N}_1)\mathbf {V}=\mathbf {\Lambda }_1\\\text{and} \quad \quad & \mathbf {V}^{(\mathbf {B}^*+\mathbf {N}_2)\mathbf {V}=\mathbf {\Lambda }_2}where \mathbf {\Lambda }_1 and \mathbf {\Lambda }_2 are positivedefinite \emph {diagonal} matrices. Let{\begin{@align}{1}{-1}\mathbf {\Lambda }_3:=\mathbf {V}^{(\mathbf {B}^*+\mathbf {N}_3)\mathbf {V}}\end{@align}be an n \times n positive definite matrix. By (\ref {eq:LVC-2}),{\begin{@align}{1}{-1}\mathbf {\Lambda }_1^{-1}+\mu \mathbf {\Lambda }_3^{-1}=(1+\mu )\mathbf {\Lambda }_2^{-1}. \end{@align}}Thus, \mathbf {\Lambda }_3 is also diagonal. Moreover, since \mathbf {N}_1\preceq \mathbf {N}_3,{\begin{@align*}{1}{-1}\mathbf {\Lambda }_1-\mathbf {\Lambda }_3 =\mathbf {V}^\mathbf {N}_1-\mathbf {N}_3)\mathbf {V}\preceq 0.\end{@align*}}and hence{\begin{@align}{1}{-1}\mathbf {\Lambda }_1 \preceq \mathbf {\Lambda }_3. \end{@align}}}\emph {Step 2--Choosing matrix parameter \mathbf {A}.} Let\tilde{\mathbf {\Lambda }}_3=\mathbf {\Lambda }_3+\epsilon \mathbf {I} for some\epsilon >0, and let \tilde{\mathbf {\Lambda }}_2 be an n \times n matrix such that{\begin{@align}{1}{-1}\mathbf {\Lambda }_1^{-1}+\mu \tilde{\mathbf {\Lambda }}_3^{-1}=(1+\mu )\tilde{\mathbf {\Lambda }}_2^{-1}. \end{@align}}Clearly, \tilde{\mathbf {\Lambda }}_2 is diagonal. Moreover, by(\ref {eq:ord1}){\begin{@align}{1}{-1}\mathbf {\Lambda }_1 \prec \tilde{\mathbf {\Lambda }}_3. \end{@align}}Note that \mu >0 so by (\ref {eq:LV-cd3}) and (\ref {eq:ord0}){\begin{@align}{1}{-1}\mathbf {\Lambda }_1 \prec \tilde{\mathbf {\Lambda }}_2 \prec \tilde{\mathbf {\Lambda }}_3. \end{@align}}Comparing (\ref {eq:LV-cd2}) and (\ref {eq:LV-cd3}) and using the factthat \mathbf {\Lambda }_3 \prec \tilde{\mathbf {\Lambda }}_3, we have{\begin{@align}{1}{-1}\mathbf {\Lambda }_2 \prec \tilde{\mathbf {\Lambda }}_2. \end{@align}}Now let{\begin{@align*}{1}{-1}& \mathbf {Y}_1 := \mathbf {V}^\mathbf {X}+\mathbf {Z}_1)\\& \mathbf {Y}_2 := \mathbf {V}^\mathbf {X}+\widetilde{\mathbf {Z}}_2)\\\mbox{and} \quad \quad & \mathbf {Y}_3 := \mathbf {V}^\mathbf {X}+\widetilde{\mathbf {Z}}_3)\end{@align*}}where \widetilde{\mathbf {Z}}_2 and \widetilde{\mathbf {Z}}_3 are Gaussian n-vectors with covariancematrices{\begin{@align*}{1}{-1}\widetilde{\mathbf {N}}_2 &= \mathbf {V}^{-\tilde{\mathbf {\Lambda }}_2\mathbf {V}^{-1}-\mathbf {B}^*\\&\succ \mathbf {V}^{-\mathbf {\Lambda }_2\mathbf {V}^{-1}-\mathbf {B}^*\\&=(\mathbf {B}^*+\mathbf {N}_2)-\mathbf {B}^*\\&=\mathbf {N}_2}}and{\begin{@align*}{1}{-1}\widetilde{\mathbf {N}}_3 &= \mathbf {V}^{-\tilde{\mathbf {\Lambda }}_3\mathbf {V}^{-1}-\mathbf {B}^*\\&= \mathbf {V}^{-(\mathbf {\Lambda }_3+\epsilon \mathbf {I})\mathbf {V}^{-1}-\mathbf {B}^*\\&=(\mathbf {B}^*+\mathbf {N}_3+\epsilon \mathbf {V}^{-\mathbf {V}^{-1})-\mathbf {B}^*\\&=\mathbf {N}_3+\epsilon \mathbf {V}^{-\mathbf {V}^{-1}}}respectively and are independent of \mathbf {X}. The covariance matrices of \mathbf {Y}_k,k=1,2,3, can be calculated as\mathbf {V}^{\sf Cov}(\mathbf {X})-\mathbf {B}^*]\mathbf {V}+\mathbf {\Lambda }_1,\mathbf {V}^{\sf Cov}(\mathbf {X})-\mathbf {B}^*]\mathbf {V}+\tilde{\mathbf {\Lambda }}_2 and\mathbf {V}^{\sf Cov}(\mathbf {X})-\mathbf {B}^*]\mathbf {V}+\tilde{\mathbf {\Lambda }}_3, respectively. Thus,\mathbf {Y}_2 and \mathbf {Y}_3 can be equivalently written as{\begin{@align*}{1}{-1}& \mathbf {Y}_3 = \mathbf {Y}_1+\mathbf {Z}\\\mbox{and} \quad \quad & \mathbf {Y}_2=\mathbf {Y}_1+\mathbf {A}^{\frac{1}{2}}\mathbf {Z}\end{@align*}}where \mathbf {Z} is a Gaussian n-vector with covariance matrix\tilde{\mathbf {\Lambda }}_3-\mathbf {\Lambda }_1\succ 0 and isindependent of \mathbf {Y}_1, and{\begin{@align}{1}{-1}\mathbf {A}&:=(\tilde{\mathbf {\Lambda }}_2-\mathbf {\Lambda }_1)(\tilde{\mathbf {\Lambda }}_3-\mathbf {\Lambda }_1)^{-1}.\end{@align}}Clearly, \mathbf {A} is diagonal. Moreover, by (\ref {eq:ord2}) 0 \prec \mathbf {A}\prec \mathbf {I}.}\emph {Step 3--Applying generalized Costa^{\prime }s EPI.} By the generalized Costa EPI(\ref {eq:VC-epi}),{\begin{@align*}{1}{-1}h(\mathbf {Y}_2|U)\ge \frac{n}{2}\log \left\lbrace |\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\exp \left[\frac{2}{n}h(\mathbf {Y}_1|U)\right]+|\mathbf {A}|^{\frac{1}{n}}\exp \left[\frac{2}{n}h(\mathbf {Y}_3|U)\right]\right\rbrace .\end{@align*}}Thus,{\begin{@align}{1}{-1}h(&\mathbf {Y}_1|U)+\mu h(\mathbf {Y}_3|U) -(1+\mu )h(\mathbf {Y}_2|U) \\&\le h(\mathbf {Y}_1|U)+\mu h(\mathbf {Y}_3|U) -\frac{(1+\mu )n}{2}\log \left\lbrace |\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\exp \left[\frac{2}{n}h(\mathbf {Y}_1|U)\right]+|\mathbf {A}|^{\frac{1}{n}}\exp \left[\frac{2}{n}h(\mathbf {Y}_3|U)\right]\right\rbrace .\end{@align}}Now we consider the function{\begin{@align*}{1}{-1}f(b,c)=b+\mu c-\frac{(1+\mu )n}{2}\log \left[|\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\exp \left(\frac{2b}{n}\right)+|\mathbf {A}|^{\frac{1}{n}}\exp \left(\frac{2c}{n}\right)\right].\end{@align*}}Note that{\begin{@align*}{1}{-1}\nabla f(b,c) &= \left[\begin{matrix} \displaystyle {1-(1+\mu )\frac{|\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\exp (2b/n)}{|\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\exp (2b/n)+|\mathbf {A}|^{\frac{1}{n}}\exp (2c/n)}}\\[3mm]\displaystyle {\mu -(1+\mu )\frac{|\mathbf {A}|^{\frac{1}{n}}\exp (2c/n)}{|\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\exp (2b/n)+|\mathbf {A}|^{\frac{1}{n}}\exp (2c/n)}}\end{matrix}\right]\end{@align*}}and{\begin{@align*}{1}{-1}\nabla ^{2}f(b,c)=-\frac{2(1+\mu )}{n}\frac{|\mathbf {A}|^{\frac{1}{n}}|\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\exp [(2b+2c)/n]}{\left[|\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\exp (2b/n)+|\mathbf {A}|^{\frac{1}{n}} \exp (2c/n)\right]^2}\left[\begin{matrix}1 & -1 \\-1 & 1\end{matrix}\right] \preceq 0.\end{@align*}}So f(b,c) is concave in (b,c). By setting \nabla f(b,c)=0, theglobal maximum is achieved when{\begin{@align*}{1}{-1}c=b+\frac{n}{2}\log \left[\mu \left(\frac{|\mathbf {I}-\mathbf {A}|}{|\mathbf {A}|}\right)^{\frac{1}{n}}\right]\end{@align*}}and the maximum is given by{\begin{@align*}{1}{-1}\frac{\mu n}{2}\log \left[\mu \left(\frac{|\mathbf {I}-\mathbf {A}|}{|\mathbf {A}|}\right)^{\frac{1}{n}}\right]-\frac{(1+\mu )n}{2}\log \left[(1+\mu )|\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\right].\end{@align*}}Hence,{\begin{@align}{1}{-1}h(\mathbf {Y}_1|U)+&\mu h(\mathbf {Y}_3|U)-(1+\mu )h(\mathbf {Y}_2|U)\\&\le \frac{\mu n}{2}\log \left[\mu \left(\frac{|\mathbf {I}-\mathbf {A}|}{|\mathbf {A}|}\right)^{\frac{1}{n}}\right]-\frac{(1+\mu )n}{2}\log \left[(1+\mu )|\mathbf {I}-\mathbf {A}|^{\frac{1}{n}}\right].\end{@align}}}\emph {Step 4--Calculating \log |\mathbf {A}| and \log |\mathbf {I}-\mathbf {A}|.} Note that(\ref {eq:LV-cd3}) can be rewritten as{\begin{@align*}{1}{-1}\mu (\mathbf {\Lambda }_1^{-1}-\tilde{\mathbf {\Lambda }}_3^{-1})=(1+\mu )(\mathbf {\Lambda }_1^{-1}-\tilde{\mathbf {\Lambda }}_2^{-1})\end{@align*}}which gives{\begin{@align}{1}{-1}\left|\frac{\tilde{\mathbf {\Lambda }}_2-\mathbf {\Lambda }_1}{\tilde{\mathbf {\Lambda }}_3-\mathbf {\Lambda }_1}\right|&=\left(\frac{\mu }{1+\mu }\right)^n\left|\frac{\tilde{\mathbf {\Lambda }}_2}{\tilde{\mathbf {\Lambda }}_3}\right|.\end{@align}}Similarly, we have{\begin{@align*}{1}{-1}(\mathbf {\Lambda }_1^{-1}-\tilde{\mathbf {\Lambda }}_3^{-1})=(1+\mu )(\tilde{\mathbf {\Lambda }}_2^{-1}-\tilde{\mathbf {\Lambda }}_3^{-1})\end{@align*}}and hence{\begin{@align}{1}{-1}\left|\frac{\tilde{\mathbf {\Lambda }}_3-\tilde{\mathbf {\Lambda }}_2}{\tilde{\mathbf {\Lambda }}_3-\mathbf {\Lambda }_1}\right|&=\left(\frac{1}{1+\mu }\right)^n\left|\frac{\tilde{\mathbf {\Lambda }}_2}{\tilde{\mathbf {\Lambda }}_1}\right|.\end{@align}}According to the definition of \mathbf {A} in (\ref {eq:def-A}),{\begin{@align}{1}{-1}\log |\mathbf {A}| &=\log \left|\frac{\tilde{\mathbf {\Lambda }}_2-\mathbf {\Lambda }_1}{\tilde{\mathbf {\Lambda }}_3-\mathbf {\Lambda }_1}\right|\\&=\log \left[\left(\frac{\mu }{1+\mu }\right)^n\left|\frac{\tilde{\mathbf {\Lambda }}_2}{\tilde{\mathbf {\Lambda }}_3}\right|\right]\end{@align}}and{\begin{@align}{1}{-1}\log |\mathbf {I}-\mathbf {A}| &=\log \left|\frac{\tilde{\mathbf {\Lambda }}_3-\tilde{\mathbf {\Lambda }}_2}{\tilde{\mathbf {\Lambda }}_3-\mathbf {\Lambda }_1}\right|\\&=\log \left[\left(\frac{1}{1+\mu }\right)^n\left|\frac{\tilde{\mathbf {\Lambda }}_2}{\mathbf {\Lambda }_1}\right|\right]\end{@align}}where (\ref {eq:equ3}) and (\ref {eq:equ4}) follow (\ref {eq:equ1}) and(\ref {eq:equ2}), respectively. Substituting (\ref {eq:equ3}) and(\ref {eq:equ4}) into (\ref {eq:h123}), we have{\begin{@align}{1}{-1}h(\mathbf {Y}_1|U)+\mu h(\mathbf {Y}_3|U)-(1+\mu )h(\mathbf {Y}_2|U)&\le \frac{1}{2}\log |\mathbf {\Lambda }_1|+\frac{\mu }{2}\log |\tilde{\mathbf {\Lambda }}_3|-\frac{1+\mu }{2}\log |\tilde{\mathbf {\Lambda }}_2|. \end{@align}}\end{@align*}\emph {Step 5--Letting \epsilon \downarrow 0.} Note that\tilde{\mathbf {\Lambda }}_3=\mathbf {\Lambda }_3+\epsilon \mathbf {I}\rightarrow \mathbf {\Lambda }_3 and\widetilde{\mathbf {N}}_3=\mathbf {N}_3+\epsilon \mathbf {V}^{-\mathbf {V}^{-1} \rightarrow \mathbf {N}_3 in thelimit as \epsilon \downarrow 0. Moreover, by (\ref {eq:LV-cd3}) wehave \tilde{\mathbf {\Lambda }}_2 \rightarrow \mathbf {\Lambda }_2 andhence{\begin{@align*}{1}{-1}\widetilde{\mathbf {N}}_2 &= \mathbf {V}^{-\tilde{\mathbf {\Lambda }}_2\mathbf {V}^{-1}-\mathbf {B}^*\\& \rightarrow \mathbf {V}^{-\mathbf {\Lambda }_2\mathbf {V}^{-1}-\mathbf {B}^*\\&=(\mathbf {B}^*+\mathbf {N}_2)-\mathbf {B}^*\\&=\mathbf {N}_2.}}Letting \epsilon \downarrow 0 on both sides of (\ref {eq:h123b}),we have{\begin{@align}{1}{-1}h(\mathbf {V}^\mathbf {X}+\mathbf {N}_1)|U)+\mu h(\mathbf {V}^\mathbf {X}+\mathbf {N}_3)|U)-&(1+\mu )h(\mathbf {V}^\mathbf {X}+\mathbf {N}_2)|U)\\&\le \frac{1}{2}\log |\mathbf {\Lambda }_1|+\frac{\mu }{2}\log |\mathbf {\Lambda }_3|-\frac{1+\mu }{2}\log |\mathbf {\Lambda }_2|. \end{@align}}Using the fact that{\begin{@align*}{1}{-1}h(\mathbf {V}^\mathbf {X}+\mathbf {N}_1)|U)=h(\mathbf {X}+\mathbf {N}_1|U)+\log |\mathbf {V}|\end{@align*}}and{\begin{@align*}{1}{-1}\log |\mathbf {\Lambda }_k|&=\log |\mathbf {V}^{(\mathbf {B}^*+\mathbf {N}_k)\mathbf {V}|\\&=\log |\mathbf {B}^*+\mathbf {N}_k|+2\log |\mathbf {V}|}\end{@align*}for k=1,2,3, the desired inequality (\ref {eq:Lepi-2}) can beobtained from (\ref {eq:h123c}). This completes the proof of thecorollary.}\section {Proof of Corollary~\ref {cor:Lepi-K}} Here, we prove Corollary~\ref {cor:Lepi-K} using mathematicalinduction. Note that when K=1, (\ref {eq:LVC-K}) implies that\mathbf {N}_1=\mathbf {N}_0. Thus, the inequality (\ref {eq:Lepi-K}) holdstrivially with equality for any (U,\mathbf {X}) independent of(\mathbf {Z}_0,\mathbf {Z}_1).\end{@align*}Assume that the inequality (\ref {eq:Lepi-K}) holds for K=Q-1. Let\mathbf {N} be an n \times n symmetric matrix such that{\begin{@align}{1}{-1}(\mathbf {B}^*+\mathbf {N})^{-1}=\sum _{k=1}^{Q-1}\mu _k^{\prime }(\mathbf {B}^*+\mathbf {N}_k)^{-1}\end{@align}}where{\begin{@align*}{1}{-1}\mu _k^{\prime }:=\frac{\mu _k}{\sum _{j=1}^{Q-1}\mu _j}, \quad j=1,\ldots ,Q.\end{@align*}}By the assumption \mathbf {N}_1 \preceq \ldots \preceq \mathbf {N}_{Q-1}, we havefrom (\ref {eq:Nt3-1}){\begin{@align}{1}{-1}\mathbf {N}_1 \preceq \mathbf {N}\preceq \mathbf {N}_{Q-1}. \end{@align}}Let \mathbf {Z} be a Gaussian random n-vector with covariance matrix\mathbf {N} and independent of (U,\mathbf {X}). By the induction assumption and(\ref {eq:Nt3-1}),{\begin{@align}{1}{-1}\sum _{k=1}^{Q-1} \mu _k^{\prime } h(\mathbf {X}+\mathbf {Z}_k|U)-h(\mathbf {X}+\mathbf {Z}|U)&\le \sum _{k=1}^{Q-1} \frac{\mu _k^{\prime }}{2} \log |\mathbf {B}+\mathbf {N}_k|-\frac{1}{2}\log |\mathbf {B}+\mathbf {N}|.\end{@align}}}On the other hand, substitute (\ref {eq:Nt3-1}) into (\ref {eq:LVC-K})and we have{\begin{@align*}{1}{-1}(\mathbf {B}+\mathbf {N})^{-1}+\mu _Q^{\prime }(\mathbf {B}+\mathbf {N}_Q)^{-1}=(1+\mu _Q^{\prime })(\mathbf {B}+\mathbf {N}_0)^{-1}.\end{@align*}}Note from (\ref {eq:ord4}) that \mathbf {N}\preceq \mathbf {N}_{Q-1} \preceq \mathbf {N}_Q. Thus, by Corollary~\ref {cor:Lepi-2}{\begin{@align}{1}{-1}h(\mathbf {X}+\mathbf {Z}|U) + \mu _Q^{\prime } h&(\mathbf {X}+\mathbf {Z}_Q|U) - (1+\mu _Q^{\prime })h(\mathbf {X}+\mathbf {Z}_0|U)\\&\le \frac{1}{2}\log |\mathbf {B}+\mathbf {N}|+\frac{\mu _Q^{\prime }}{2}\log |\mathbf {B}+\mathbf {N}_Q|-\frac{1+\mu _Q^{\prime }}{2}\log |\mathbf {B}+\mathbf {N}_0|.\end{@align}}Putting together (\ref {eq:Qv-3}) and (\ref {eq:Qv-4}), we have{\begin{@align*}{1}{-1}\sum _{j=1}^{Q} \mu _j h(\mathbf {X}+\mathbf {Z}_j|U)- h(\mathbf {X}+\mathbf {Z}_0|U)&\le \sum _{j=1}^{Q} \frac{\mu _j}{2} \log |\mathbf {B}+\mathbf {N}_j|-\frac{1}{2}\log |\mathbf {B}+\mathbf {N}_0|.\end{@align*}}This proved the induction step and hence the corollary.}\bibliographystyle {IEEEtran}\begin{thebibliography}{10}\url@samestyle \end{thebibliography}\bibitem {Sha-BSTJ48}C.~E. Shannon, ``A mathematical theory of communication,^{\prime \prime } \emph {Bell Syst.Tech. 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